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3 years ago

How many protons, neutrons, and electrons does bi-210 have

Physics

1 answer:

nikdorinn [45]3 years ago

5 0

Answer:

Explanation:

The number of neutrons is 127 and the number of protons will be equal to number of electrons. It has 83 protons and a magic number of 127 neutrons. Bi 210 is radioactive isotope of bismuth.

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A 1900kg airplane is flying at an altitude of 510 above the ground. What is the gravitational potential energy in Joules?

Natalija [7]

Answer: 9496200 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the moving plane since it moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 1900kg

g = 9.8m/s^2

h = 510 metres (units of height is metres)

Thus, GPE = 1900kg x 9.8m/s^2 x 510m

GPE = 9496200 joules

Thus, the gravitational potential energy of the airplane is 9496200 joules

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3 years ago

A scientist working with Polonium-210 detects radiation when the nucleus decays. The radiation emitted has very little penetrati

Anton [14]

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3 years ago

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What is the term aerodynamic?

Neporo4naja [7]

Aerodynamics is the way air moves around things. The rules of aerodynamics explain how an airplane is able to fly. Anything that moves through air reacts to aerodynamics.

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3 years ago

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A 110 kg tackler moving at 2.5m/s meets head on (and holds on to) an 82 kg halfback moving at 5.0m/s. what will be their mutual

Leona [35]

Answer:

0.7 m/s

Explanation:

We can solve the problem by using conservation of momentum.

Before the collision, the momentum of the first tackler is:

$p_1 = m_1 v_1 =(110 kg)(2.5 m/s)=275 kg m/s$

while the momentum of the second tackler is

$p_2 = m_2 v_2 =(82 kg)(-5.0 m/s)=-410 kg m/s$

Note that we used a negative sign because the direction of the second tackler is opposite to that of the first tackler.

Therefore, the total momentum before the collision is

$p_i = p_1 +p_2 =275 kg m/s -410 kg m/s=-135 kg m/s$

Since the total momentum is conserved, this is also equal to the final total momentum :

$p_f = -135 kg m/s$

Which is also equal to

$p_f = (m_1 +m_2 )v_f$

since the two tacklers continue their motion together with final velocity vf. Re-arranging the previous equation, we can find the the new velocity of the two tacklers:

$v_f = \frac{p_f}{m_1 +m_2}=\frac{-135 kg m/s}{110 kg + 82 kg}=-0.7 m/s$

and the negative sign means the direction is the one of the second tackler.

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3 years ago

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A 5 cm radius isolated conducting sphere is charged so its potential is 100 V, relative to the potential far away. The charge de

MArishka [77]

The electric potential (V) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation:

$V = \frac{kQ}{r}$